## Question 15.81

$\frac{d[R]}{dt}=-k[R]^{2}; \frac{1}{[R]}=kt + \frac{1}{[R]_{0}}; t_{\frac{1}{2}}=\frac{1}{k[R]_{0}}$

104822659
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### Question 15.81

Hi, I'm not understanding the loss and reverse part of the question. If anyone can help, that would be great!
The equilibrium constant for the second-order attachment of a substrate to the active site of an enzyme was found to be 326 at 310 K. At the same temperature, the rate constant for the second-order attachment is 7.4 x10^ 7 L*mol^-1*s^-1. What is the rate constant for the loss of unreacted substrate from the active site (the reverse of the attachment reaction)?

Chem_Mod
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### Re: Question 15.81

The question is really just asking the relationship between forward reaction rate constant k, reverse reaction rate constant k' and the equilibrium constant K.